2

Actually that is, I would like to use a complex exponent to turn a number x into an -x just by applying some complex exponent to x. I think I would have to use some ln, π and so one, but I am actually stack and cannot find the correct path to solving it.

For example, let's say for 2 we choose the real part to be equal 1, what would be the solution for y in this special case.

$$ 2^{1 + iy}=-2 $$

What would be the solution for the imaginary part of z if the real part must be 0.5?

$$ x^{0.5 + iy} = -x $$

$$ x^{0.5} * x^{iy} = \sqrt{x} * -\sqrt{x} = -x $$

What is y in this case?

Is there maybe a general solution for

$$ x^{z}=-x $$

webdeb
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  • http://www.wolframalpha.com/input/?i=solve+e%5E(ixa)%3D-a – Bobson Dugnutt Mar 12 '17 at 17:33
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    Unless $x$ happens to be a positive real number, this equation $x^z = -x$ is not well-defined for $z$ other than integers. Of course it will be fairly rare that $x$ allows an integer exponent solution. – hardmath Mar 12 '17 at 18:00
  • Actually I just would like to understand it for $$2^z=-2$$ and $$3^z=-3$$ Specially in the case where $$ real part of (z) = 0.5 $$ – webdeb Mar 12 '17 at 18:04

1 Answers1

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Let $f(x)=1+\frac{i\pi}{\ln(x)}$ for $x\neq 0$, and $f(x)=1$ for $x=0$. Then $x^{f(x)}=-x$ for all $x\in\Re_0^+$.

This is obvious for $x=0$. For $x\neq 0$ we have: $-x=x(-1) = x^{1+\log_x (-1)} = x ^{1+\frac{\ln(-1)}{\ln(x)}} = x ^{1+\frac{i\pi}{\ln(x)}}$.

Anonymous
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    That is clearly not what OP means. – Bobson Dugnutt Mar 12 '17 at 17:41
  • Is there a solution like this, where the real part is 0.5 ? – webdeb Mar 12 '17 at 17:45
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    @Lovsovs it may be clear to you, but it's certainly not clear to me. The OP is asking for $f(x)$ such that $x^{f(x)}=-1$ for all $x$. Now, $f(x)=1+\frac{i\pi}{\ln(x)}$ satisfies the requirement. What's wrong with that? – Anonymous Mar 12 '17 at 17:46
  • What if $x$ is not real? – Bernard Mar 12 '17 at 17:53
  • A lot of editing has been going on since then, I have no problem with your answer now! – Bobson Dugnutt Mar 12 '17 at 18:01
  • Indeed there are more solutions of the type $x ^{1+\frac{(2k+1)\pi}{\ln(x)}i}$ for integer $k$ (assuming $x$ is a positive real) – Henry Mar 12 '17 at 18:05
  • hmm, ok it plays well if real part is 1, @Anonymous is there a nice solution if real part is 0.5? – webdeb Mar 12 '17 at 18:08
  • @webdeb not for all x! This is obvious e.g. for $x=e$, since $e^{1/2+iy}$ with $y\in\Re$ has modulus $\sqrt{e}\neq e$. For some x (say, $0$ :)) it certainly works. – Anonymous Mar 12 '17 at 18:14
  • Ok, I just would like to know what the solution would be for x = 2 or 3 – webdeb Mar 12 '17 at 18:15
  • @webdeb Well, for $x=2$ you have $f(x)=1+\frac{i\pi}{\ln(2)}$. In fact, as Henry pointed out, for any integer k, $f_k(x)=1+\frac{(2k+1)i\pi}{\ln(2)}$ will work. But if you are working with a positive real $x$, the real part of $f(x)$ can't be chosen arbitrarily (say, $0.5$) if you want $x^{f(x)}$ and $x$ to have the same modulus. – Anonymous Mar 12 '17 at 18:24
  • Oh, no the real part is 1, is it doable when the real part is 1/2? – webdeb Mar 12 '17 at 18:28
  • I mean is there a way to calculate yi if the real part is 1/2 – webdeb Mar 12 '17 at 18:33
  • it would mea the x^yi would result in $$-\sqrt{x} $$ – webdeb Mar 12 '17 at 18:35
  • @webdeb AH! That's something completely different! You should open a different question. – Anonymous Mar 12 '17 at 18:37
  • Yeah, it makes sense I did: http://math.stackexchange.com/questions/2183668/complex-exponent-with-real-part-1-2-should-convert-x-into-x – webdeb Mar 12 '17 at 18:45